Question

What is the ph after 0.195 mol of naoh is added to the buffer from part a? assume no volume change on the addition of the base?

Answer 1

When 0.195 mole of NaOH are added and there is no volume change, moles of base will increase by 0.195 and moles of acid will decrease by 0.195 mole
[Base] = 0.608 + 0.195 = 0.803 mole
[Acid] = 0.809 - 0.195 = 0.614 mole
pH = 6.25 + log $\frac{0.803}{0.614} = 6.37$ 

Answer 2

pH after addition of NaOH = 6.364

Further explanation

A buffer solution is a solution that can maintain a good pH value due to the addition of a little acid or a little base or dilution.

The buffer solution can be acidic or basic

• Acid buffer solutions consist of weak acids and their salts.

$\displaystyle [H ^ +] = Ka \times \frac {mole \: weak \: acid} {mole \: salt \times valence}$

valence according to the amount of salt anion

• Alkaline buffer solution consists of a weak base and its salt.

$\displaystyle [OH ^ -] = Kb \times \frac {mole \: weak \: base} {mole \: salt \times valence}$

valence according to the amount of salt cation

• If the buffer solution is acidic then

a slight addition of acid (H⁺) will be balanced by the conjugate base

the addition of a small base (OH⁻) will be balanced by the weak acid

• If the buffer solution is alkaline then

the addition of a little acid (H⁺) will be balanced weak base

the addition of a small base (OH⁻) will be balanced

the weak acid HA and NaA, including Acid buffer solution

the addition of a small base (OH⁻) will be balanced by the weak acid

Part A: What is the pH of a buffer prepared by adding 0.809 moles of the weak acid HA to 0.609 moles of NaAin 2.00 L of solution? The dissociation constant Ka of HA is 5.66×10−7

The ph after 0.195 mol of naoh is added to the buffer from part A:

Reactions that occur:

HA      + NaOH ---> NaA + H2O

0.809     0.195        0.609

0.195       0.195        0.195

0.614       0               0.804

$\displaystyle [H^+]=Ka \times \frac {mole\: weak\:acid}{mole \:salt \times valence}$

$\displaystyle [H^+]=5.66.10^{-7} \times \frac {0.614}{0.804\times 1}$

$\displaystyle pH=-log~4.322.10^{-7}\\\\pH=6.364$

Learn more

pH scale

brainly.com/question/599764

brainly.com/question/11063271

solution has the lowest ph

brainly.com/question/9875355

Keywords : buffer, acid, salt, pH